Q.12 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 12 Solution

Topic: Properties of Bulk Matter · Official NTA Booklet Code 50

Question

In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity (v) with the ratio of density of spherical ball (σ) to density of the liquid (ρ), is best represented by:

✓

Verified Against NTA Final Answer Key

Option (1) : Graph (1): Line starts at sigma / rho = 1 on the horizontal axis and increases linearly

📚 Properties of Bulk Matter Medium

📋 Given Parameters & Problem Setup

Experiment Measurement of viscosity of liquids using terminal velocity

Spherical Balls Identical radius r, varying density sigma

Liquid Density rho (constant)

Variables to Plot Terminal velocity v vs density ratio (sigma / rho)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Root Finding & Physical Cutoff Shortcut

⏱️ Bypasses complex algebraic plotting.

Find where terminal velocity equals zero!

1 Recall Stokes' Law terminal velocity: v_t = (2/9) * [r^2 * g / eta] * (sigma - rho).

2 Factor out liquid density rho: v_t = C * [(sigma / rho) - 1], where C = (2/9) * r^2 * g * rho / eta is positive.

3 Set v_t = 0 ==> (sigma / rho) - 1 = 0 ==> sigma / rho = 1.

4 Therefore, the graph MUST intersect the horizontal axis at exactly sigma / rho = 1.

5 For sigma / rho < 1, the sphere is less dense than the liquid, so buoyant force exceeds gravity; the ball floats and cannot fall downward.

6 Hence, v = 0 for sigma/rho <= 1, and increases linearly with slope C for sigma/rho > 1.

7 Look at the graphs: Only Graph (1) starts directly at (1, 0) on the axis. Solved in 15 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Terminal velocity represents steady-state descent where Weight = Buoyant Force + Viscous Drag. At sigma = rho, Weight = Buoyancy, so net sinking force is zero and v_t = 0.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulate Dynamic Equilibrium for a Falling Sphere

When a sphere reaches terminal velocity v_t in a viscous medium, net downward acceleration is zero:

F_{\text{net}} = W - F_B - F_v = 0 \implies m g - \rho V g - 6\pi \eta r v_t = 0

• Weight of sphere: W = sigma * V * g = sigma * (4/3 pi r^3) * g.
• Upthrust (Buoyant force): F_B = rho * V * g = rho * (4/3 pi r^3) * g.
• Stokes' viscous drag force: F_v = 6 pi eta r v_t.

Step 2

Express Terminal Velocity in terms of Density Ratio

Equate net gravitational force to viscous drag and solve for v_t:

v_t = \frac{2}{9} \frac{r^2 g}{\eta} (\sigma - \rho) = \left( \frac{2 r^2 g \rho}{9 \eta} \right) \left( \frac{\sigma}{\rho} - 1 \right)

• Let x = sigma / rho and constant k = (2 r^2 g rho) / (9 eta).
• Then v_t = k * (x - 1).
• This is a straight line equation of the form y = m(x - x_0) with x-intercept x_0 = 1.

Step 3

Analyze Physical Boundary Conditions and Graph Shape

Evaluate physical behavior across different density regimes:

v_t = 0 \quad \text{for} \quad \frac{\sigma}{\rho} \le 1, \qquad v_t \propto \left(\frac{\sigma}{\rho} - 1\right) \quad \text{for} \quad \frac{\sigma}{\rho} > 1

• If sigma < rho, the ball floats upward (or remains at rest on the surface); falling terminal velocity is 0.
• At sigma / rho = 1, downward terminal velocity is exactly 0.
• For sigma / rho > 1, terminal velocity increases linearly with slope k.
• Matches Graph (1) perfectly, corresponding to Option (1).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 Option (1)

✓ 100% Correct

Verdict & Scientific Reason:

Exact physical representation of Stokes' Law and Archimedes' principle; confirmed by Official NTA Key.

💡

Why this is airtight: Correctly identifies that terminal descent requires sigma > rho and starts at x = 1.

Note: Flawless experimental physics.

2 Option (2)

✗ Incorrect

Verdict & Scientific Reason:

Violates static equilibrium.

⚠️

Student Fallacy (Why candidates pick this trap): Assumes the ball would already fall with terminal speed even when its density equals liquid density.

Note: When sigma = rho, buoyancy completely balances weight; terminal sinking speed must be zero.

3 Option (3)

✗ Incorrect

Verdict & Scientific Reason:

Neglects buoyant upthrust completely.

⚠️

Student Fallacy (Why candidates pick this trap): Plots v_t proportional to sigma without subtracting liquid density rho.

Note: Buoyant force reduces net driving weight: net force is proportional to (sigma - rho), not sigma alone.

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Ignores dependence of driving weight on ball density.

⚠️

Student Fallacy (Why candidates pick this trap): Assumes terminal velocity is a fixed constant independent of sphere mass.

Note: Heavier spheres of the same size sink faster through viscous drag.

⚠️

NTA Trap Alert: Neglecting Buoyant Upthrust (The Origin Extrapolation Trap)

Students who forget Archimedes' principle assume v_t ~ sigma and pick Graph (3), which passes through the origin.

🛡️ How to avoid the -1 mark: Always remember: In any fluid motion, effective weight is ALWAYS (sigma - rho)*V*g. The driving term is (sigma - rho), so velocity is zero when sigma = rho.

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Core NEET Syllabus Standard

NEET UG Physics Core Standards · Mechanical Properties of Fluids Standard Benchmark

Core Scientific Principle Tested:

"Terminal velocity of a spherical body falling in a fluid: v_t = (2/9)*(r^2*g/eta)*(sigma - rho). Sinking occurs only when sigma > rho."

Syllabus Focus: Stokes' Law, Terminal Velocity, and Practical Measurement of Viscosity

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 12 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Root Finding & Physical Cutoff Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Formulate Dynamic Equilibrium for a Falling Sphere

Express Terminal Velocity in terms of Density Ratio

Analyze Physical Boundary Conditions and Graph Shape

360° Option Elimination Engine Key Feature

NTA Trap Alert: Neglecting Buoyant Upthrust (The Origin Extrapolation Trap)

Core NEET Syllabus Standard

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